Not stationary conditions

  • Thread starter caldus2311
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  • #1
Hello.

I would like to find an analytical solution of the one group point kinetic equations with the external source. I have a critical reactor and I would do the step change in reactivity. I would like to find the solution for n(t).

The ONE-GROUP point kinetics equations:

dn/dt=([tex]\rho[/tex]-[tex]\beta[/tex])/[tex]\Lambda[/tex]*n(t)+[tex]\lambda[/tex]*c(t)+Q

dc/dt=[tex]\beta[/tex]/[tex]\Lambda[/tex]*n(t)-[tex]\lambda[/tex]*c(t)

I can solve the equations using Laplace transformations, but I don`t know the initial conditions: c(0) and n(0), since the reactor is critical I can`t use the stationary initial conditions?

What can I do.

Thanks.
 
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Answers and Replies

  • #2
Astronuc
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since the reactor is critical I can`t use the stationary initial conditions?
One uses the steady-state initial considitions, n(0)=n0 and c(0) =c0. The initial flux has to be proportional to the power in the system. Remember the system can be critical at zero power, at full rated power, or at any state in between. If the system exceeds its full rated power, the reactor protection system would scram the reactor.

There has to be some equilibrium value of any fission product, including delayed neutron precursors.
 
  • #3
Thanks, Astronuc.

Maybe I didn`t quoted my problem very well in my previous post.

In a subcritical reactor (k<1) you can say, that before you will do a step change in reactivity you have a stationary state, where dn/dt=0 and dc/dt=0. So you can express c(0) with n(0). In a one group approximation it follows c(0)=[tex]\beta[/tex]*n(0)/([tex]\lambda[/tex]*[tex]\Lambda[/tex]) and n(0)=-Q*[tex]\Lambda[/tex]/[tex]\rho[/tex]0,

where [tex]\rho[/tex]0 is the reactivity before the step change and Q is the external source of neutrons.

But in a critical reactor (k>1) you can`t say that dn/dt=0 and dc/dt=0, since the power is rising. I am looking the expressions for c(0) and n(0) in a critical reactor.


Thanks.
 
  • #4
Astronuc
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In a critical reactor, k=1, and the power is constant.

In a supercritical reactor, k > 1, and the power increases, and in a subcritical system, k < 1, and the power decreases. But somewhere before reactivity is added or removed, the system is critical.


For k = 1, n(t) is constant, and the n(0) is proportional to power. Knowing n(0), one obtains c(0), which it appears one did.

In a one group approximation it follows c(0)=[tex]\beta[/tex]*n(0)/([tex]\lambda[/tex]*[tex]\Lambda[/tex]) and n(0)=-Q*[tex]\Lambda[/tex]/[tex]\rho[/tex]0,

where [tex]\rho[/tex]0 is the reactivity before the step change and Q is the external source of neutrons.
But is Q necessarily an external source, or is it just a source - including internal as well as external?
 
Last edited:
  • #5
Astronuc: I understand your last post.

OK. From the begining.

Let say that I have a subcritical reactor, with the reactivity [tex]\rho[/tex]=-1,5. Than I make a step change in reactivity to [tex]\rho[/tex]=-1. In this example I can calculate the n(t) from the equations from my first post. Before the step change in reactivity I can say that dn(t=0)/dt=0. The system is in steady state. The number of neutrons does not change.
So I can calculate n(0)!

BUT, when I have a reactor in a state, when k>1, for example [tex]\rho[/tex]=0.002 and I hypothetically increase the reactivity to [tex]\rho[/tex]=0.003. In this case I can`t say that dn(t=0)/dt=0!!.

My question. What can I do in this last case to calculate n(t) when I change the reactivity from 0.002[tex]\rightarrow[/tex]0.003???

Yes, Q is the external source of neutrons.

Thanks.
 
  • #6
Astronuc
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Taking the example of a transition from [itex]\rho[/itex] from 0.002 to 0.003, then one would in principle start with k = 1 before insertion of reactivity of 0.002, in which case, one has n(t) for some t after the reactivity has changed. Then adding additional reactivity of 0.001 for a total of 0.003 is solvable, but using n(t) and c(t) at the time that the additional reactivity is added.

Such would be the case when one withdraws a control rod/blade a few steps, then withdraws a second set of rods/blades or the same set a few more steps before the reactor reaches steady-state again.

When one inserts reactivity and the power increases, the negative reactivity also increases via Doppler effect (increased resonance broadening with increased fuel temperature) and reduction in moderator coefficient (density of moderator decreases with temperature).
 

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